Question: Let $g(x)=x^4+4x^3-18x^2$. For what values of $x$ does the graph of $g$ have a point of inflection? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-4$ (Choice B) B $x=-3$ (Choice C) C $x=1$ (Choice D) D $g$ has no points of inflection.
Answer: We can find the inflection points of the graph of $g$ by looking for the intervals where its second derivative $g''$ is positive/negative. This analysis is very similar to finding minimum/maximum points, only instead of analyzing $g'$, we are analyzing $g''$. The second derivative of $g$ is $g''(x)=12(x+3)(x-1)$. $g''(x)=0$ for $x=-3,1$. Since $g''$ is a polynomial, it's defined for all real numbers. Therefore, our possible inflection points are $x=-3$ and $x=1$. Our possible inflection points divide the number line into three intervals: $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $(-\infty, \llap{-}3)$ $( \llap{-}3,1)$ $(1,\infty)$ Let's evaluate $g''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g''(x)$ Verdict $(-\infty,-3)$ $x=-4$ $g''(-4)=60>0$ $g$ is concave up $\cup$ $(-3,1)$ $x=0$ $g''(0)=-36<0$ $g$ is concave down $\cap$ $(1,\infty)$ $x=2$ $g''(2)=60>0$ $g$ is concave up $\cup$ We can see that the graph of $g$ changes concavity at both $x=-3$ and $x=1$. In conclusion, these are the values of $x$ where the graph of $g$ has a point of inflection: $x=-3$ $x=1$